# Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products HCHO_(2(aq))+H_2O_((l))⇌H_3O_((aq))^(+)+CHO_(2(aq))^(-). Use A for [HCHO_2], B for [H_2O], C for [H_3O^+], D for [CHO_2^(−)]?

##### 1 Answer
May 23, 2015

The equilibrium constant expresses the ratio the exists between the concentrations of the products and the concentration of the reactants, each raised to the power of their stoichiometric coefficient, at equilibrium.

For a general equilibrium reaction, you can write the equilibrium constant like this

$a A + b B r i g h t \le f t h a r p \infty n s c C + \mathrm{dD}$

${K}_{c} = \frac{{\left[C\right]}^{c} \cdot {\left[D\right]}^{d}}{{\left[A\right]}^{a} \cdot {\left[B\right]}^{b}}$

Now, when you're in aqueous solution, the concentration of the water is not included in the expression of the equilibrium constant because it is assumed constant.

In other words, the concentration of the water is so large compared with the concentrations of the other species that take part in the reaction that any change brought to it by the reaction can be ignored.

In your case, the equilibrium reaction looks like this

$H C H {O}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + C H {O}_{2 \left(a q\right)}^{-}$

If you rewrite this using the letters assigned to each compound, you'll get

${A}_{\left(a q\right)} + {B}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {C}_{\left(a q\right)} + {D}_{\left(a q\right)}$

This means that ${K}_{c}$ will be equal to

${K}_{c} = \frac{\left[C\right] \cdot \left[D\right]}{\left[A\right]}$

Notice that, in your case, all the stoichiometric coefficients are equal to 1.